Find the Intersection of 2 Sets in Unioned Interval Notation

Mathematica, 41 bytes - 15% = 34.85

Mathematica has a built-in function for interval intersection.

List@@IntervalIntersection@@Interval@@@#&

Example:

In[1]:= List@@IntervalIntersection@@Interval@@@#&[{{{-90, -4}, {4, Infinity}}, {{-50,Infinity}}}]

Out[1]= {{-50, -4}, {4, Infinity}}

Haskell, 145 bytes

import Data.List
q(a,b)=[a,a+0.5..b]
p@(a,b)%(h:t)|h==b+0.5=(a,h)%t|1<2=p:(h,h)%t
p%_=[p]
a#b|h:t<-nub$sort$intersect(q=<<a)$q=<<b=(h,h)%t|1<2=[]

Usage example: [(-2.0,0.0),(2.0,4.0),(5.0,6.0),(6.0,8.0)] # [(-1.0,1.0),(3.0,5.0),(5.0,9.0)] -> [(-1.0,0.0),(3.0,4.0),(5.0,8.0)].

How it works:

                 q=<<a            -- turn each pair in the 1st input list into
                                  -- lists with halves in between (e.g. (1,4) ->
                                  -- [1,1.5,2,2.5,3,3.5,4]) and concatenate them
                                  -- to a single list
                      q=<<b       -- same for the second input list
    nub$sort$intersect            -- sort the intersection of those lists
                                  -- and remove duplicates
h:t<-                             -- call the first element h and the rest t
                       (h,h)%t    -- start rebuilding the intervals
                          |1<2=[] -- if there's no first element h, one of the
                                  -- input lists is empty, so the output is also
                                  -- empty


p@(a,b)%(h:t)                     -- an interval p = (a,b), build from a list (h:t)
             =(a,h)%t             -- becomes (a,h)
      |h==b+1                     --   if h equals b+0.5
                    p:(h,h)%t     -- or is put in the output list, followed by
                                  --       a new interval starting with (h,h)
      |1<2                        --   otherwise
p%_=[p]                           -- base case of the interval rebuilding function 

I'm putting the "half"-values x.5 in the list, because I need to distinguish (1,2),(3,4) from (1,4). Without x.5, both would become [1,2,3,4], but with x.5 the 1st becomes [1,1.5,2,3,3.5,4] (which lacks 2.5) and the second [1,1.5,2,2.5,3,3.5,4].

APL (Dyalog Extended), 37 bytes

.5×(⊢/,⊃)¨{⍵⊂⍨1,1<2-/⍵}⊃∩/∨⍤∊¨…/¨¨2×⎕

Try it online!

A full program which takes a nested array in STDIN and returns a nested array.

Code fixed by bubbler.